Use Taylor's Theorem to prove that
$\displaystyle cosx = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}$
Thanks!
The Taylor series for the function $\displaystyle f(x)$, centred at a = 0, is given by: $\displaystyle f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x)$
where $\displaystyle R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} $ is the remainder term ($\displaystyle c \in (0,x)$). To prove that the Taylor series of $\displaystyle f(x)$ converges to $\displaystyle f(x)$ for all $\displaystyle x$, we must show that the interval of convergence is $\displaystyle \infty$ and $\displaystyle \lim_{n \to \infty} R_n (x) = 0$.
Coefficients will be for you to find. Use the ratio test to find the interval of convergence.
For the remainder term, notice that: $\displaystyle 0 < |R_n (x)| = \left| \frac{\cos^{(n+1)} (c)}{(n+1)!}x^{n+1}\right| {\color{red}\ < \ } \frac{x^{n+1}}{(n+1)!}$
Use squeeze theorem to finish off (recall: $\displaystyle R_n (x) \rightarrow 0 $ iff $\displaystyle |R_n (x)| \rightarrow 0)$)